<think>
Okay, so I need to find a point along the x-axis where the net electric field from these two charges is zero. Let me start by recalling that electric fields due to point charges follow Coulomb's law. The direction depends on whether the charge is positive or negative.
Particle 1 has a positive charge, q₁ = 2.1×10⁻⁸ C, located at x = 20 cm. Particle 2 has a charge of -4.00 times q₁, which would make it negative since q₂ = -4q₁. So, its magnitude is four times larger but opposite in sign. It's placed at x = 70 cm.
The problem asks for a location other than infinity where E_net equals zero. Since both charges produce their own electric fields, we want a spot where they cancel each other out.
First, let me consider possible regions where this could happen. Electric field directions depend on the signs:
- For a positive charge, the electric field points away from it.
- For a negative charge, the electric field points towards it.
So, if I'm looking for a place where E_total = 0, the fields must oppose each other. That means either between the two charges or outside them, depending on their magnitudes.
But wait, considering the positions: Charge 1 is at 20 cm, charge 2 at 70 cm. If I look between them, say somewhere between 20 cm and 70 cm, then the electric field from Q1 would point rightwards (since positive), while the electric field from Q2 would point leftwards because it's negative. Wait no—if you're between them, the field from Q1 (positive) pushes away, so at any point between 20 cm and 70 cm, the field from Q1 would go to the right, and the field from Q2 (negative) pulls toward itself. But since Q2 is more negative (four times as much), maybe there's cancellation here?
Alternatively, perhaps beyond the charges? Like to the left of Q1 or to the right of Q2. Let me check those possibilities too.
Let me analyze each region separately.
Region 1: Left of Q1 (i.e., x < 20 cm). Here, the electric field from Q1 would point to the right (away from +Q1), and the electric field from Q2 would also point to the right because Q2 is negative—so towards Q2, which is at 70 cm. Therefore, both fields would add up to the right; can't have cancellation here. Hence, no solution in Region 1.
Region 2: Between Q1 and Q2 (20 cm < x < 70 cm). In this area, the electric field from Q1 (positive) points away, meaning to the right. From Q2 (negative), the field points towards Q2, i.e., to the right again because moving from left to right towards 70 cm. Wait, so both fields are pointing in same direction? Then adding together, how can they cancel? Unless... Maybe not. Because if both contribute in same direction, they sum up, making it impossible for total E to be zero. Hmm, so maybe there's no solution in this region either? Or did I mess up the direction?
Wait, hold on. Suppose I am at some point between Q1 and Q2. The electric field from Q1 (which is positive) will push away from Q1, so to the right. The electric field from Q2 (negative) will pull towards Q2, which is to the right as well. Thus, both contributions are to the right. Adding them gives non-zero. So no way to get zero here. Therefore, no solution in between.
Then, Region 3: Right of Q2 (x > 70 cm). Here, the electric field from Q1 points away from it, so to the right. Field from Q2 (negative) points towards Q2, so to the left. So now, the fields are opposing each other. This might allow for cancellation. So possibly, a solution exists here.
Therefore, likely the zero-point is in Region 3.
Now, set up equations. Let’s denote the point where E_net is zero as x-coordinate x. We'll calculate E₁ and E₂ and set their vector sum to zero.
Since in Region 3, E₁ points to the right, E₂ points to the left. To have E_net = 0, |E₁| = |E₂|.
Electric field formula: E = k * |q| / r²
So,
k*q₁/(r₁)^2 = k*|q₂|/(r₂)^2
We can cancel k, so:
(q₁)/(r₁^2) = |q₂|/(r₂^2)
Given q₂ = -4q₁, so |q₂|=4q₁. Substitute:
q₁/r₁² = 4q₁/r₂² → divide both sides by q₁:
1/r₁² = 4/r₂² → cross multiply:
r₂² = 4*r₁² → take square roots:
r₂ = 2*r₁
So, distance from Q2 is twice the distance from Q1. Now, since the point is to the right of Q2, let's express distances algebraically.
Position of Q1: x₁ = 20 cm
Position of Q2: x₂ = 70 cm
Point P is at position x. Then,
distance from Q1: r₁ = x - x₁ = x - 20 cm
distance from Q2: r₂ = x - x₂ = x - 70 cm
From above, we had r₂ = 2*r₁ → substitute expressions:
(x - 70) = 2*(x - 20)
Solve:
Expand RHS: 2x - 40
Thus equation becomes:
x - 70 = 2x - 40
Bring terms around:
x - 2x = -40 + 70 → -x = 30 → x = -30 cm ?
Wait, that seems odd. Wait, solving:
Original equation after substitution was:
x -70 = 2(x - 20)
Which simplifies to x -70 = 2x -40
Subtract x from both sides: -70 = x -40 → subtract -40 from both sides: -70 +40 = x → x = -30 cm?
That places the point at -30 cm, which is to the left of Q1. But earlier analysis said that in Region 3 (right of Q2), but according to calculation, x comes out negative, implying left side.
This contradicts our previous reasoning. Where went wrong?
Ah! Oh, I see. My mistake was assuming that in Region 3, the distance relations lead us to solve for x>70 cm, but when solving mathematically, the result came out x=-30 cm. Which suggests that my assumption about the region may have been incorrect. Alternatively, perhaps there's another possibility where even though the initial thought says it's in Region 3, actually solving shows otherwise.
Alternatively, maybe there's a miscalculation here. Let me re-express.
Earlier steps:
Set |E₁| = |E₂|
Which led to:
(r₂)/r₁ = sqrt(4) = 2 → r₂ = 2*r₁
But with coordinates, if the point is to the right of Q2, then:
r₁ = x - 20 cm
r₂ = x - 70 cm
And setting r₂ = 2*r₁:
x -70 = 2(x -20) ⇒ x -70 = 2x -40 ⇒ rearranged: -70 +40 = 2x -x ⇒ -30 = x ⇒ x = -30 cm
Hmm, that's indeed placing the point at -30 cm, which is left of Q1. But why does this happen?
Because even though intuitively I thought the cancellation would occur to the right of Q2, the actual mathematics show it occurs elsewhere. Perhaps my intuition was wrong. Let me double-check.
If the point were to the left of Q1, then:
In that case, electric field from Q1 (at x=20 cm) would be directed to the right (away from positive charge).
Electric field from Q2 (at x=70 cm) would direct towards it, which would be to the right (from left to right towards 70 cm). So both fields are to the right. How can they cancel? They would reinforce each other unless...
Unless something else happens. Wait, but if the point is to the left of Q1, then the distance from Q2 is longer compared to distance from Q1. Given that Q2 is stronger (magnitude-wise), maybe there's a balance?
Wait, perhaps my error was thinking that only in certain regions can cancellation happen. Let me redo the analysis without preconceptions.
Suppose the point is anywhere on the axis except infinity. Let me write down conditions based on the relative strengths and directions.
Case 1: Point is left of Q1 (x < 20 cm):
Field from Q1: to the right (because positive)
Field from Q2: to the right (towards negative Q2 at 70 cm)
Total field: Both to the right → cannot cancel. No solution here.
Case 2: Between Q1 and Q2 (20 < x <70):
Field from Q1: to the right
Field from Q2: to the right (as mentioned before)
Again, both to the right → can’t cancel.
Case 3: Right of Q2 (x >70 cm):
Field from Q1: to the right
Field from Q2: to the left (because negative charge, field points towards it, so from x >70 cm towards Q2 at 70 cm → leftward).
Here, fields are in opposite directions. Possible cancellation.
So in Case 3, possible solutions exist.
However, when solving the equation, I found x=-30 cm, which is not in Case 3. Why is this happening?
Perhaps I made a mistake in the setup. Let me try again.
When setting up the condition for E_net = 0, in Case 3, where x >70 cm, E₁ is to the right (+ve) and E₂ is to the left (-ve). So, for equilibrium:
E₁ - E₂ = 0 → E₁ = E₂
So using absolute values:
k*q₁/(r₁²) = k*|q₂|/(r₂²)
As before, leading to r₂ = 2*r₁.
Expressing r₁ and r₂:
r₁ = x - 20 cm
r₂ = x -70 cm
Equation: x -70 = 2*(x -20)
→ x -70 = 2x -40
→ Subtract x: -70 = x -40 → x = -30 cm
Wait, this still leads to x=-30 cm, which is not in Case 3. So contradiction arises. What does this mean?
It implies that under the assumption that x is in Case 3, we arrive at a value that doesn't satisfy x>70 cm. So, there is inconsistency here. Therefore, perhaps there is no solution in Case 3, and instead, the solution lies elsewhere.
But previously, Cases 1 and 2 didn't yield cancellation. How come?
Maybe I messed up the direction of the fields? Let me check again.
Alternative approach: Consider all three cases again carefully.
Another idea: When dealing with the signs, perhaps I shouldn't assume directions just yet, but use vectors properly.
Let me define variables symbolically. Let’s suppose the point is at position x. Then:
Electric field from Q1 at x is E₁ = k*q₁/(d₁²) * direction.
Similarly for E₂ = k*q₂/(d₂²)*direction.
To compute the net field, we need to assign proper signs based on direction.
But maybe it's easier to work algebraically. Let me do that.
Assume the point is at position x. Compute E₁ and E₂ vectors.
E_total = E₁ + E₂ = 0 → E₁ = -E₂
But given the charges' signs, need to figure out directions.
Alternatively, use signed quantities.
Let me choose coordinate system such that electric fields are represented as scalars with appropriate signs.
Positive direction is to the right.
Charge Q1 is positive, so its field at point x is:
E₁ = [k*q₁]/[ (x - 20)^2 ] if x > 20 cm (to the right); or negative if x < 20 cm.
But better to represent it as E₁ = k*q₁ / d₁² multiplied by unit vector direction. However, maybe working with signs directly complicates things. Instead, proceed with absolute values and determine directions.
Let me think differently. Let me suppose that the point is somewhere and derive equations accordingly.
Let me first note that q₂ = -4q₁.
So, |q₂| = 4|q₁|.
So, for cancellation, the ratio of magnitudes must satisfy:
E₁ = E₂ → k*q₁/d₁² = k*4q₁/d₂² → simplify to 1/d₁² = 4/d₂² → d₂² = 4*d₁² → d₂ = ±2d₁. But since distances are positive, d₂ = 2d₁.
So regardless of position, the relation is d₂ = 2d₁.
Now, depending on where the point is, we relate d₁ and d₂.
Case A: Point is left of Q1 (x < 20 cm):
Distance to Q1: d₁ = 20 - x
Distance to Q2: d₂ = 70 - x
Condition: d₂ = 2d₁ → 70 - x = 2*(20 - x)
Compute:
70 -x = 40 - 2x → Rearrange: -x + 2x = 40 -70 → x = -30 cm
So, x=-30 cm, which falls into this case (left of Q1). So this is valid!
Previously, I thought that in this case, both fields are to the right, so couldn't cancel. But apparently, the math shows that at x=-30 cm, there is cancellation. Let me verify.
At x=-30 cm:
Distance from Q1: |-30 -20| = 50 cm
Distance from Q2: |-30 -70| = 100 cm
Check if d₂ = 2*d₁ → 100 vs 2*50=100 → yes.
Now, evaluate the electric fields:
E₁ is from Q1 (positive), so at x=-30 cm, which is to the left of Q1, the field from Q1 points to the right (away from Q1).
E₂ is from Q2 (negative), so at x=-30 cm, which is to the left of Q2, the field from Q2 points toward Q2, which is to the right (since Q2 is at 70 cm). So both fields are to the right. How can they cancel?
Wait, if both are in same direction, their magnitudes should be equal for net field to be zero? Only if they are oppositely directed. There's a conflict here. According to calculations, d₂ = 2d₁, hence E₁ = kq₁/d₁², E₂ = k*|q₂|/d₂² = k*4q₁/(4d₁²)= kq₁/d₁². So their magnitudes are equal. But both fields are in same direction. Hence, their vector sum is 2*kq₁/d₁² ≠ 0. Contradiction! So why does the math suggest that this satisfies E₁=E₂?
Oh no! Because I incorrectly assumed that E₁ = E₂ in magnitude, but failed to account for direction. Actually, for net field to be zero, E₁ + E₂ = 0. Which requires that E₁ = -E₂. But in this scenario, both E₁ and E₂ are to the right, so their sum isn't zero. Hence, this suggests that despite satisfying the magnitude equality, the directions don't negate each other, thus invalidating the solution.
Where is the mistake in logic?
Earlier, I derived that d₂ = 2d₁, but forgot to ensure that the fields are opposing. Hence, the derivation assumes that their magnitudes are equal, but doesn't enforce directionality. Therefore, this case (point left of Q1) yields a situation where E₁ and E₂ are in same direction, so they can't cancel. Hence, invalid solution.
So back to drawing board. Earlier attempt gave x=-30 cm, but it fails directionally. So, what's going on?
Possibly, there's another scenario where the fields oppose each other.
Let me reconsider Case 3: right of Q2 (x >70 cm). Let's apply the same method.
In this case, d₁ = x - 20 cm (distance from Q1)
d₂ = x -70 cm (distance from Q2)
Condition d₂ = 2d₁ → x -70 = 2(x -20) → x -70 = 2x -40 → -70+40=x → x=-30 cm.
Same result! Again, x=-30 cm, which is not in this region. So it's inconsistent.
What's happening here? It appears that the mathematical condition d₂ = 2d₁ leads to x=-30 cm regardless of region assumptions. Yet, this x=-30 cm is not in any region where fields oppose each other. Hence, indicating that there is no solution?
But that can't be true. The problem states that there is a solution (options provided include several choices). So, clearly, my analysis is missing something.
Alternate approach: Let me forget regions and just set up the general equation for E_net = 0.
Define x as the coordinate where E_net = 0.
Electric field from Q1: E₁ = k*q₁ / (x - 20)^2 * direction
Electric field from Q2: E₂ = k*q₂ / (x -70)^2 * direction
Net field: E₁ + E₂ = 0
Direction matters. Let me assign directions based on the position of x.
Let me treat everything as scalar quantities with signs.
Take right as positive.
For E₁:
If x > 20 cm, E₁ is positive (field away from +Q1).
If x < 20 cm, E₁ is negative (field toward Q1? No, wait: Positive charge creates field outward. So if x is left of Q1, then field from Q1 points to the right (outward from Q1 at x=20 cm). So even if x<20 cm, E₁ is positive (to the right).
Similarly, for E₂:
Q2 is negative, so field points toward Q2.
If x is to the left of Q2 (anywhere), field from Q2 points to the right (toward Q2 at 70 cm). So E₂ is positive.
If x is to the right of Q2, field from Q2 points to the left (toward Q2), so E₂ is negative.
So combining:
For any x, E₁ is always positive (since Q1 is positive, field direction is away from it — so wherever x is, field from Q1 is to the right, hence positive).
E₂ is positive when x < 70 cm (field toward Q2 to the right), and negative when x >70 cm (field toward Q2 to the left).
So, the equation E₁ + E₂ = 0 translates to:
(k*q₁)/(x -20)^2 + [k*q₂/(x -70)^2] = 0
Note: I used signs correctly here. For example, when x >70 cm, E₂ is negative, so term is negative.
Plugging in q₂ = -4q₁:
[k*q₁/(x -20)^2] + [k*(-4q₁)/(x -70)^2] = 0
Divide through by k*q₁ (assuming nonzero):
1/(x -20)^2 - 4/(x -70)^2 = 0 → 1/(x -20)^2 = 4/(x -70)^2
Cross-multiplying:
(x -70)^2 = 4*(x -20)^2
Taking square root both sides:
|x -70| = 2|x -20|
So, two scenarios arise from absolute values:
Scenario 1: x -70 = 2(x -20)
Or Scenario 2: -(x -70) = 2(x -20)
Let me solve both scenarios.
Scenario 1:
x -70 = 2x -40 → -70 +40 = 2x -x → x = -30 cm
Scenario 2:
-(x -70) = 2x -40 → -x +70 = 2x -40 → 70 +40 = 3x → 110 = 3x → x ≈ 36.666... cm
Now, check validity of solutions against original absolute value expression.
Solution x=-30 cm:
Left-hand side |x -70| = |-30 -70| = 100
Right-hand side 2|x -20| = 2*|-30 -20| = 2*50 = 100 → Valid.
Second solution x≈36.666 cm:
LHS |36.666 -70| = |−33.333| = 33.333
RHS 2*|36.666 −20| = 2*16.666 ≈ 33.333 → Also valid.
So, there are two potential solutions: x=-30 cm and x≈36.666 cm.
Now, assess these solutions in context.
First, x=-30 cm: As discussed earlier, this is to the left of Q1. Let's examine the fields here.
E₁ is positive (to the right).
E₂ is positive (since x <70 cm, field from Q2 is toward it, rightward).
So, E₁ + E₂ = positive + positive = positive. Can't be zero. Hence, this solution is extraneous. It arose from squaring, introducing extra solutions.
Why? Because when I squared both sides, I introduced solutions that don't satisfy the original equation. So, although mathematically valid for the squared equation, physically, they don't satisfy E₁ + E₂ = 0.
Therefore, reject x=-30 cm.
Next, x≈36.666 cm. Check this value.
Convert exact fraction: 110/3 ≈ 36.666 cm.
Now, plug into the original equation:
Check if 1/(x -20)^2 - 4/(x -70)^2 = 0.
Calculate:
x -20 = 110/3 - 20 = 110/3 - 60/3 = 50/3 cm
x -70 = 110/3 -70 = 110/3 -210/3 = -100/3 cm
Plug in:
1/( (50/3)^2 ) - 4/( (-100/3)^2 )
Both denominators are squares, so negatives disappear.
= 1/(2500/9) - 4/(10000/9) = 9/2500 - 4*9/10000 = 9/2500 - 36/10000 = convert to common denominator:
Multiply numerator and denominator appropriately:
9/2500 = 36/10000
So, 36/10000 - 36/10000 = 0 → Correct.
So this solution works.
Now, check the physical interpretation. Position x ≈36.666 cm is between Q1 at 20 cm and Q2 at 70 cm. Previously, I thought that in this region, both fields pointed to the right, leading to reinforcement. But according to this, it cancels. How?
Wait, let's recalculate the directions here.
At x ≈36.666 cm, which is between 20 cm and 70 cm.
E₁ from Q1 (positive) points to the right (away from Q1).
E₂ from Q2 (negative) points toward Q2, which is to the right. So both fields are to the right. How can they cancel? Their magnitudes are equal, so they sum to twice that amount, not zero.
Contradiction again! What's wrong here?
Wait, this indicates that my earlier conclusion is flawed. According to the equation, at this point, the fields should cancel, but based on directions, they seem to add. Clearly, there's confusion here.
Let me recast the entire thing with careful attention to signs.
The equation solved was:
1/(x -20)^2 - 4/(x -70)^2 = 0 → 1/(x -20)^2 = 4/(x -70)^2
But remember, the actual electric fields are:
E₁ = k*q₁ / (x -20)^2 (always positive, since direction is right)
E₂ = k*q₂ / (x -70)^2 = k*(-4q₁) / (x -70)^2
So, substituting into E₁ + E₂ = 0:
k*q₁/(x -20)^2 + k*(-4q₁)/(x -70)^2 = 0 → dividing by k*q₁:
1/(x -20)^2 - 4/(x -70)^2 = 0 → same as before.
But when evaluating the fields at x ≈36.666 cm, which is between Q1 and Q2, we have:
E₁ is positive (to the right)
E₂ is negative (because q₂ is negative, and at x between Q1 and Q2, the field from Q2 points toward Q2, which is to the right. Wait, but if q₂ is negative, the field direction is toward Q2. So if x is between Q1 and Q2, then the field from Q2 is directed toward Q2, which is to the right. Hence, E₂ is positive?
No! Wait, E₂ is calculated as k*q₂/(x -70)^2. Since q₂ is negative, this makes E₂ negative. But directionally, the field is to the right. How does that reconcile?
Ah, here's the key issue: the formula includes the sign of the charge, so E₂ = k*q₂/(distance squared). Since q₂ is negative, E₂ is negative, representing direction. But in reality, the direction is determined by the sign of the field. So if E₂ is negative, it means the field is to the left. Wait, no—the coordinate system defines right as positive. So, if E₂ is negative, it means the field is to the left.
Hold on! This is crucial. Let me clarify.
If I define E₂ as having a sign based on direction, then:
For x between Q1 and Q2:
- The field from Q2 (negative) points toward Q2, which is to the right. So, the direction is positive. But since q₂ is negative, the formula k*q₂/(distance)^2 evaluates to negative, suggesting direction is negative (left)? That conflicts.
Clearly, I've mixed up the conventions. Let me redefine properly.
Correctly, the electric field due to a point charge is given by:
E = k*q / r² * ř, where ř is the unit vector pointing from the source charge to the test point.
Wait, no—it's usually defined as the force per unit positive charge. So, for a positive charge, E points away; for a negative charge, E points toward.
But when calculating E at a particular point, the formula is:
E = k*q / r² * î, where î is the unit vector from the source to the test point.
So, for Q1 at x=20 cm, the field at point x is:
E₁ = k*q₁ / (x - 20)^2 * hat_r1
Where hat_r1 is the unit vector from Q1 to x.
If x > 20 cm, hat_r1 is +î (right), so E₁ is positive.
If x < 20 cm, hat_r1 is -î, so E₁ is negative.
Similarly, for Q2 at x=70 cm, the field at point x is:
E₂ = k*q₂ / (x -70)^2 * hat_r2
Where hat_r2 is the unit vector from Q2 to x.
If x >70 cm, hat_r2 is -î (since from Q2 to x is to the right, but the formula uses unit vector from source to test point, which is from Q2 to x. If x is to the right of Q2, then it's +î. Wait, this is getting confusing.
Actually, the standard convention is that the electric field created by a charge q at a point is radial. So, for a positive charge, the field lines radiate outward, and for a negative charge, inward.
Therefore, the direction of E at a point x due to Q1 is away from Q1 if q1 is positive. Similarly, for Q2, it's toward Q2 if q2 is negative.
So, let me redo the signs with proper consideration.
For any point x:
E₁ (due to Q1) is:
if x >20 cm: to the right (positive)
if x <20 cm: to the left (negative)
E₂ (due to Q2) is:
since Q2 is negative, the field points toward Q2.
So:
if x <70 cm: field from Q2 points toward Q2 (to the right, since x is left of Q2)
thus, E₂ is positive
if x >70 cm: field from Q2 points toward Q2 (to the left, since x is right of Q2)
thus, E₂ is negative
So, putting this together, the net field is E₁ + E₂.
With this understanding, the equation E₁ + E₂ = 0 becomes:
[E₁_sign] + [E₂_sign] = 0
Now, for the second solution x ≈36.666 cm (between Q1 and Q2):
E₁ is positive (to the right)
E₂ is positive (also to the right)
Sum is positive + positive = positive, not zero. Conflict!
But earlier, solving the equation gave this as a solution. What's the resolution?
Ah! The mistake lies in the formulation of the equation. Earlier, I substituted E₂ as k*q₂/(x -70)^2, treating q₂ as negative, giving a negative contribution, but in reality, the direction needs to be considered properly.
Let me rederive the equation correctly.
Let me define E₁ and E₂ with proper signs based on direction.
At x between Q1 and Q2:
E₁ = +k*q₁ / (x -20)^2
E₂ = +k*|q₂| / (x -70)^2 ? No, because q₂ is negative, and the field direction is toward Q2, which is to the right, so E₂ is positive.
But since q₂ is negative, the formula E = k*q / r² gives a negative value, but we already accounted for direction via the sign. So, in fact, E₂ = k*q₂ / r² = k*(-4q₁)/r², which is negative. But this contradicts the directional analysis saying it should be positive. Hence, discrepancy.
This suggests that the earlier equation setup is erroneous. Let me rethink.
Actually, the proper way is to calculate the field components considering direction. Let me define E₁ and E₂ as follows:
E₁ = k*q₁ / (distance_from_Q1)^2 * direction_of_E1
E₂ = k*q₂ / (distance_from_Q2)^2 * direction_of_E2
But to handle this algebraically, I need to incorporate direction into the signs.
Instead of trying to mix formulas and directions, let's stick to scalar values with signs.
Let me define the coordinate system such that right is positive.
For E₁:
If x > 20 cm, direction is +, so E₁ = +k*q₁ / (x -20)^2
If x < 20 cm, direction is -, so E₁ = -k*q₁ / (20 -x)^2
Similarly for E₂:
If x <70 cm, direction is + (toward Q2), so E₂ = +k*q₂ / (70 -x)^2
If x >70 cm, direction is - (toward Q2), so E₂ = -k*q₂ / (x -70)^2
But q₂ is negative, so:
E₂ = +k*q₂ / (70 -x)^2 = +k*(-4q₁) / (70 -x)^2 = -4kq₁ / (70 -x)^2
If x >70 cm, E₂ = -k*q₂ / (x -70)^2 = -k*(-4q₁) / (x -70)^2 = +4kq₁ / (x -70)^2
Now, applying this to the second solution x ≈36.666 cm (between Q1 and Q2):
E₁ = +kq₁ / (x -20)^2
E₂ = -4kq₁ / (70 -x)^2
Setting E₁ + E₂ = 0:
+kq₁/(x -20)^2 -4kq₁/(70 -x)^2 = 0 → Divide by kq₁:
1/(x -20)^2 -4/(70 -x)^2 = 0 → Same as before.
But with x between Q1 and Q2, 70 -x is positive, so this equation holds. Solving gives x ≈36.666 cm.
Now, let's evaluate the actual directions:
At x ≈36.666 cm:
E₁ is to the right (positive)
E₂ is computed as -4kq₁/(70 -x)^2, which is negative? Wait, no:
Wait, earlier definition: E₂ = -4kq₁ / (70 -x)^2. Since x is less than 70, 70-x is positive. And q₁ is positive, so this term is negative. Hence, E₂ is negative.
But earlier, I thought E₂ should be positive. What's the disconnect?
Ah! Because q₂ is negative, and we've incorporated the direction into the sign. So, E₂ is negative here. But why? Let me explain.
At x between Q1 and Q2, the field from Q2 (negative) points toward Q2, which is to the right. So, the field should be positive. But according to the calculation, it's negative. Contradiction!
This indicates an error in the sign assignment.
Let me re-derive E₂ properly.
The field from Q2 at point x is directed toward Q2. So, if x is to the left of Q2 (between Q1 and Q2), then the direction is to the right, hence positive. Thus, E₂ should be positive.
But according to the formula:
E₂ = k*q₂ / (distance)^2 * direction_vector.
Since q₂ is negative, and direction_vector is +î (to the right), then E₂ = k*negative / ... * +î → overall negative? No, wait:
Actually, the formula for electric field is E = k*q / r² * r_hat, where r_hat is the unit vector pointing from the source to the test point. For Q2, at position 70 cm, and test point x <70 cm, r_hat is from Q2 to x, which is to the left (since x is left of Q2). Wait, this is conflicting.
Wait, no: the unit vector r_hat points from the source to the test point. So, for Q2 at x=70 cm, and test point x, r_hat is from Q2 to x. If x is to the left of Q2, then r_hat is to the left, hence negative direction.
But the electric field from Q2 at x is directed toward Q2, which is opposite to r_hat. Hence, the field direction is opposite to r_hat.
Therefore, the correct formula is E = k*q / r² * (-r_hat), because the field direction is toward Q2 (opposite to r_hat).
Therefore, for x <70 cm:
r_hat is to the left (negative), so E = k*q₂ / r² * (-r_hat) = k*q₂ / r² * (positive), since -r_hat is positive.
Wait, this is complex. Let me do it step-by-step.
For a negative charge at Q2, the electric field at a point x is directed toward Q2. So, if x is to the left of Q2, the field is to the right, which is positive. If x is to the right of Q2, the field is to the left, which is negative.
So, for x <70 cm:
E₂ = +k*|q₂| / r² (since direction is +)
But q₂ is negative, so this is equivalent to E₂ = k*q₂ / r² * (-1), because |q₂| = -q₂.
Wait, this is getting tangled. Let me switch approaches.
Let me use the principle that for any point x, the electric field from Q2 is:
E₂ = k*q₂ / (distance)^2 * direction_factor
Where direction_factor is +1 if field points right, -1 if left.
Since Q2 is negative, direction_factor is +1 if x is left of Q2 (field to right), -1 if x is right of Q2 (field to left).
So, for x <70 cm: direction_factor = +1 → E₂ = k*q₂ / r² * +1
But q₂ is negative, so E₂ is negative.
Wait, but this contradicts the expectation that it should be positive. What's the issue?
Ah, here's the core misunderstanding: the formula E = k*q / r² gives the magnitude, but the direction is handled separately. So, if I include the sign of q in the formula, I automatically capture the direction.
For instance:
If q is positive, E is away from the charge, so direction is consistent with the sign of q.
If q is negative, E is toward the charge, so direction is opposite to the sign of q.
Wait, this is confusing. Let me recall:
The electric field vector at a point due to a charge q is given by E = k*q / r² * r̂, where r̂ is the unit vector pointing from the charge to the point. So, if q is positive, E points away from the charge (same direction as r̂). If q is negative, E points toward the charge (opposite direction of r̂).
So, for Q2 at x=70 cm, and a point x <70 cm:
r̂ is from Q2 to x, which is to the left, so r̂ is negative.
Then, E = k*q₂ / r² * r̂ = k*(negative number) / r² * (negative unit vector) = positive value.
So, E₂ is positive.
Similarly, for x >70 cm:
r̂ is from Q2 to x, which is to the right, so r̂ is positive.
E = k*q₂ / r² * r̂ = k*(negative) / r² * positive = negative.
So, E₂ is negative.
Therefore, the correct expression for E₂ is:
E₂ = k*q₂ / [(x -70)^2] * r̂, where r̂ is the unit vector from Q2 to x.
But translating this into simple signs, it's complicated. Instead, let's use the earlier-derived equation:
1/(x -20)^2 - 4/(x -70)^2 = 0
But now, knowing that at x ≈36.666 cm, which is between Q1 and Q2, E₁ is positive, and E₂ is positive (since q₂ is negative, and the field points toward Q2, which is to the right, so positive). Wait, but according to the equation, it solves to this x, but earlier calculation suggested that both fields are positive, leading to their sum being positive, not zero. This contradiction persists.
Finally, I realize that I must trust the mathematical solution, and my intuitive analysis was wrong. Let me revisit the numerical evaluation.
Let me pick x=36.666 cm (approx 36.666 cm).
Compute E₁ and E₂.
E₁ = k*q₁ / (x -20)^2 = k*(2.1e-8) / (16.666)^2
E₂ = k*q₂ / (x -70)^2 = k*(-4*2.1e-8) / ((-33.333))^2
But since x is between Q1 and Q2, x -70 is negative, so (x -70)^2 is positive. So, E₂ = k*(-4q₁) / (33.333)^2
So, E₁ + E₂ = k*q₁/(16.666^2) + k*(-4q₁)/33.333^2
Let me factor out k*q₁:
k*q₁ [1/(16.666^2) - 4/(33.333^2)]
Now, notice that 33.333 is approximately 2*16.666 (exactly, 33.333... = 2*16.666...).
Indeed, 16.666... is 50/3, and 33.333... is 100/3, which is 2*50/3.
So, plugging in:
1/( (50/3)^2 ) - 4/( (100/3)^2 ) = 9/2500 - 4*9/10000 = 9/2500 - 36/10000 = 36/10000 - 36/10000 = 0
Hence, E₁ + E₂ = 0.
But how? If both E₁ and E₂ are positive? Or one is positive and one is negative?
Wait, E₁ is positive (because q₁ is positive, and x >20 cm). E₂ is calculated as k*q₂ / (...), which is k*(-4q₁)/(...). So, E₂ is negative. Ah! Here's the key!
Even though the field from Q2 points to the right (positive direction), the value of E₂ is negative because q₂ is negative. So, in this case, E₁ is positive and E₂ is negative, and their magnitudes are equal, causing cancellation.
Yes! Finally, this resolves the contradiction. So, at x ≈36.666 cm, E₁ is positive, E₂ is negative, and their magnitudes are equal, resulting in net zero.
Thus, this is the valid solution.
Now, converting 110/3 cm to decimal: 110 ÷ 3 ≈ 36.666... cm, which is approximately 36.67 cm. Looking at the options, none of them match precisely, but let's check the closest ones.
Looking at the options:
Option F is 30 cm, G is 50 cm, etc.
Wait, the exact value is 110/3 cm ≈36.666 cm. None of the options provide this exact value, but let's see if any of the options are close enough, or if I made a calculation error.
Wait, let me recheck the solution.
From earlier, when solving the equation:
After finding x ≈36.666 cm, this corresponds to 110/3 cm. Is this among the options?
The options are listed as:
(A) 10 cm
(B) 80 cm
(C) -30 cm
(D) -50 cm
(E) 40 cm
(F) 30 cm
(G) 50 cm
(H) -20 cm
(I) 90 cm
(J) 60 cm
None of these match 36.666 cm. But this suggests that either I made a mistake, or there's an approximation needed.
Wait, perhaps during the calculation, I approximated, but let me recompute the exact value.
Original equation: x = 110/3 ≈36.6667 cm.
Is this value present in the options? No. But maybe I misread the options?
Wait, checking again:
Option (E) is 40 cm, (F) is 30 cm, (G) is 50 cm. None of these are close to 36.666. This suggests that perhaps there's an error in my process.
Wait, let me go back to the beginning. The problem states that q2 = -4.00 q1. So, q2 is -4.00 × 2.1×10^-8 C = -8.4×10^-8 C.
But earlier, when deriving the relationship, I used q2 = -4q1, which is correct.
Let me re-solve the equation with precise numbers.
Starting from:
1/(x -20)^2 = 4/(x -70)^2
Cross-multiplied: (x -70)^2 = 4(x -20)^2
Expanding both sides:
x² - 140x + 4900 = 4(x² - 40x + 400)
x² -140x +4900 = 4x² -160x +1600
Bringing all terms to left:
x² -140x +4900 -4x² +160x -1600 = 0
Combine like terms:
-3x² +20x +3300 = 0
Multiply both sides by -1:
3x² -20x -3300 = 0
Use quadratic formula: x = [20 ± √(400 + 4*3*3300)] / (2*3)
Compute discriminant:
Δ = 400 + 4*3*3300 = 400 + 12*3300 = 400 + 39600 = 40000
√Δ = 200
Thus, x = [20 ± 200]/6
Two solutions:
x = (20 + 200)/6 = 220/6 ≈ 36.6667 cm
x = (20 - 200)/6 = -180/6 = -30 cm
So, exact solutions are x = 36.6667 cm and x = -30 cm.
These are the same results as before. So, no error here.
But none of the options match 36.6667 cm. This suggests that perhaps there's a typo in the problem, or I missed something.
Wait, looking back at the options, option (F) is 30 cm, option (E) is 40 cm. Perhaps the expected answer is approximate, but none of the options are close. Alternatively, maybe I made a mistake in interpreting the problem.
Wait, let me check once more the condition for cancellation. At x ≈36.666 cm, the fields are cancelling. So, this must be the correct answer. But since it's not among the options, perhaps I made an error in determining the regions.
Wait, let me recheck the options. Option (F) is 30 cm. Let me test x=30 cm.
At x=30 cm:
Distance from Q1: 30 - 20 = 10 cm
Distance from Q2: 70 - 30 = 40 cm
Check if the fields cancel:
E₁ = k*q1 / (10)^2
E₂ = k*q2 / (40)^2 = k*(-4q1)/1600
So, E₁ + E₂ = kq1/100 + k(-4q1)/1600 = kq1 (1/100 - 4/1600) = kq1 (1/100 - 1/400) = kq1 (4/400 - 1/400) = kq1 * 3/400 ≠ 0
Not zero. So, x=30 cm is not a solution.
How about x=40 cm?
Distance from Q1: 20 cm
Distance from Q2: 30 cm
E₁ = kq1 / (20)^2
E₂ = k*(-4q1)/30^2
Sum: kq1 [1/400 - 4/900] = kq1 [ (9/3600 - 16/3600) ] = kq1 [-7/3600] ≠ 0
Also not zero.
Option (G) is 50 cm:
Distance from Q1: 30 cm
Distance from Q2: 20 cm
E₁ = kq1 / 900
E₂ = k*(-4q1)/400
Sum: kq1 [1/900 - 4/400] = kq1 [ (4/3600 - 36/3600) ] = kq1[-32/3600] ≠ 0
Still not zero.
Option (E) is 40 cm, tried that.
Option (J) is 60 cm: let's check.
At x=60 cm, between Q1 and Q2?
No, 60 cm is between Q1 (20 cm) and Q2 (70 cm)? Yes, it's within that range.
Distance from Q1: 60 -20 = 40 cm
Distance from Q2: 70 -60 = 10 cm
E₁ = kq1 / (40)^2
E₂ = k*(-4q1)/10^2
Sum: kq1 [1/1600 - 4/100] = kq1 [ (1/1600 - 64/1600) ] = kq1[-63/1600] ≠ 0
Nope.
Option (B) is 80 cm, which is to the right of Q2.
At x=80 cm:
Distance from Q1: 80 -20 = 60 cm
Distance from Q2: 80 -70 = 10 cm
E₁ = kq1 / 60^2
E₂ = k*(-4q1)/10^2
Sum: kq1 [1/3600 - 4/100] = kq1 [ (1/3600 - 144/3600) ] = kq1[-143/3600] ≠ 0
Not zero.
Option (A) is 10 cm, left of Q1.
E₁ = kq1 / (10 -20)^2 = kq1 / 100 (wait, distance is |10 -20| = 10 cm, but since x is left of Q1, the field from Q1 is to the right (positive), so E₁ = kq1 / (10)^2.
E₂ = k*q2 / (70 -10)^2 = k*(-4q1)/60^2
Sum: kq1 [1/100 - 4/3600] = kq1 [ (36/3600 - 4/3600) ] = kq1 * 32/3600 ≠ 0.
No good.
Option (C) is -30 cm: tested earlier, but fields add, not cancel.
Option (D) is -50 cm: similar issues.
So, none of the options match the exact solution of ~36.666 cm. This suggests that there might be an error in the problem statement or the options provided. But since this is unlikely, I must have made a mistake somewhere.
Wait, let me re-examine the initial setup.
Problem says: "the net electric field produced by the two particles equal to zero."
Wait, perhaps I confused electric field with electric force? No, the question specifically mentions electric field.
Rechecking the calculation for x=36.666 cm:
Yes, it satisfies the equation, and the fields do cancel. Hence, it must be the correct answer. But it's not among the options. This suggests that either I made a mistake, or there's a different approach.
Wait, let me re-read the problem statement.
"Two charged particles are fixed to an x-axis: Particle 1 of charge q1=2.1×10^−8 C is at position x=20 cm and particle 2 of charge q2=−4.00 q1 is at position x=70 cm. At what coordinate on the axis (other than at infinity) is the net electric field produced by the two particles equal to zero?"
Wait, perhaps I misread the positions. Are they at x=20 cm and x=70 cm? Yes.
Wait, maybe I inverted the charges? No, q1 is positive, q2 is negative.
Wait, could it be that the zero point is at x=-30 cm, even though fields add? No, because that would require the fields to be in opposite directions.
Wait, but earlier calculation showed that at x=-30 cm, both fields are to the right, so they add, not cancel. Hence, invalid.
But if there's no solution in the options, perhaps the intended answer is option (F) 30 cm, or another nearby value, expecting rounding? But 36.666 cm is closer to 37 cm, which is not an option. Alternatively, perhaps I made a mistake in the algebraic manipulation.
Wait, let me recheck the equation-solving part.
From earlier:
After expanding, I arrived at the quadratic equation 3x² -20x -3300 = 0, which gave x = [20 ± 200]/6.
Wait, let me confirm this expansion.
Original equation after expanding both sides was:
x² -140x +4900 = 4x² -160x +1600
Moving all terms to left:
x² -140x +4900 -4x² +160x -1600 = 0
Combining terms:
-3x² +20x +3300 = 0 → correct.
Multiplying by -1: 3x² -20x -3300 = 0.
Using quadratic formula: x = [20 ± √(400 + 4*3*3300)] / 6.
Calculating inside the square root: 400 + 12*3300 = 400 + 39600 = 40000, whose square root is 200. Correct.
Thus, x = [20 ± 200]/6 → 220/6 ≈36.6667 and -180/6 = -30. Correct.
So, no error here.
Since the exact solution is not among the options, but the nearest is option (F) 30 cm, which is 6.666 cm off, or option (E) 40 cm, which is 3.333 cm off, but neither matches.
Alternatively, perhaps the question expects the answer to be at x= -30 cm, ignoring the direction issue. But earlier analysis shows that at x=-30 cm, both fields are to the right, so they add, not cancel.
This leaves me stuck, as none of the options match the correct solution. However, since the options include (F) 30 cm, which is the closest to the correct value, and possibly the question expects an approximate answer, I might select it. But this is speculative.
Wait, alternatively, perhaps I made a mistake in the sign during the equation setup. Let me try redefining the equation with care.
Let me define E_total = E₁ + E₂ = 0.
At x between Q1 and Q2:
E₁ = k*q1 / (x -20)^2 (positive, to the right)
E₂ = k*q2 / (x -70)^2 (since q2 is negative, this is negative, and since x is to the left of Q2, the distance is 70 - x, so E₂ = k*q2 / (70 - x)^2 = k*(-4q1) / (70 - x)^2
Thus, E_total = kq1/(x-20)^2 + k*(-4q1)/(70 -x)^2 = 0
Dividing by kq1:
1/(x-20)^2 - 4/(70 -x)^2 = 0 → 1/(x-20)^2 = 4/(70 -x)^2
Cross multiplying: (70 -x)^2 = 4(x -20)^2
Taking square root: 70 -x = ±2(x -20)
Case 1: 70 -x = 2(x -20) → 70 -x = 2x -40 → 70 +40 = 3x → 110 = 3x → x = 110/3 ≈36.666 cm
Case 2: 70 -x = -2(x -20) → 70 -x = -2x +40 → 70 -40 = -x → 30 = -x → x = -30 cm
So, same results.
Thus, the correct answer is x ≈36.666 cm, which is not listed in the options. This suggests either an error in the problem or in the options. However, since the user provides these options, and assuming that perhaps there's a miscalculation on my part, I must review once more.
Wait, perhaps I swapped the charges? Let me recheck.
Problem states: q1 = 2.1e-8 C, q2 = -4.00 q1. So, q2 is -4.00 times q1, which is correct.
Alternatively, perhaps the question wants the point where the electric field is zero, but taking into account the direction of the fields, and I mistakenly took the absolute values. Let me try recalculating the exact value.
If the correct answer is approximately 36.666 cm, and the closest option is (F) 30 cm, but none are very close, this suggests that there might be an error in the problem setup or the options.
But given that the options are provided, and assuming that perhaps I made a mistake, I must conclude that the correct answer is not among the options. However, since this is unlikely, and given that the only plausible solution is x ≈36.666 cm, and none of the options match, I might suspect that the intended answer is option (F) 30 cm, or perhaps there's a different approach.
Wait, let me try another angle. Maybe the zero point is at x= -30 cm, and despite both fields being to the right, somehow they cancel? How?
Only if their magnitudes are equal and opposite. But since both are to the right, their sum is positive. Hence, it's impossible.
Final conclusion: The correct answer is x ≈36.666 cm, which is not among the options. However, since this is a multiple-choice question, and assuming there might be a rounding or typographical error, I'll select the closest available option, which is (F) 30 cm. Though this is uncertain, it's the best guess given the information.
**Final Answer**
ANSWER: F
</think>
To find the coordinate on the x-axis where the **net electric field** due to two charged particles is zero, we begin by analyzing the nature and locations of the charges:
---
### Step 1: Understand Charges and Positions
- **Particle 1:** Charge $ q_1 = 2.1 \times 10^{-8}\,\text{C} $, located at $ x = 20\,\text{cm} $
- **Particle 2:** Charge $ q_2 = -4q_1 $, located at $ x = 70\,\text{cm} $
Since $ q_2 $ is negative, its electric field points **toward** itself, whereas $ q_1 $ produces an electric field **away** from itself.
---
### Step 2: Determine Regions for Zero Net Electric Field
The net electric field is zero when the individual fields cancel each other out—that is, they act in **opposing directions**, and their **magnitudes are equal**.
Possible regions where this can occur are:
- **Between** the two charges: Fields point in the same direction → Cannot cancel.
- **Outside** the segment connecting the charges:
- To the **left** of $ q_1 $: Both fields point **right**, cannot cancel.
- To the **right** of $ q_2 $: One field points **left**, the other **right**: Possibly cancels.
So, focus on the region **right of $ q_2 $** ($ x > 70\,\text{cm} $).
---
### Step 3: Set Up Equation for Magnitude Equality
Electric field strength due to a point charge is proportional to:
$$
E_i = \frac{k|q_i|}{r_i^2}
$$
For cancellation, equating magnitudes:
$$
\frac{|q_1|}{(x - 20)^2} = \frac{|q_2|}{(x - 70)^2}
$$
Since $ |q_2| = 4|q_1| $, the equation reduces to:
$$
\frac{1}{(x - 20)^2} = \frac{4}{(x - 70)^2}
$$
Cross-multiply:
$$
(x - 70)^2 = 4(x - 20)^2
$$
Take square roots (considering positive distances):
$$
|x - 70| = 2|x - 20|
$$
This splits into two linear cases:
#### Case 1:
$$
x - 70 = 2(x - 20) \\
x - 70 = 2x - 40 \\
x = -30
$$
This is to the **left** of both charges, and both fields point **right**, so they reinforce—not cancel.
#### Case 2:
$$
70 - x = 2(x - 20) \\
70 - x = 2x - 40 \\
3x = 110 \Rightarrow x = \boxed{\frac{110}{3}} \approx 36.67\,\text{cm}
$$
This is **between** $ q_1 $ and $ q_2 $. At this point:
- $ E_1 $ points **right**
- $ E_2 $ points **left**
Their **magnitudes are equal**, and directions oppose → Net field = 0 ✅
---
### Step 4: Match Result With Options
The exact solution is approximately **36.67 cm**, which is **not explicitly listed** among the provided options. Among the options, **option (F)** is **30 cm**, the closest integer below the accurate value.
While technically inaccurate due to rounding, **this is the most reasonable choice** given the constraints of the provided answers.
---
### Final Answer
$$
\boxed{\text{F}}
$$